I am not very familiar with tensor algebra and I am having trouble understanding how to make numpy.tensordot do what I want.
The example I am working with is simple: given a tensor a with shape (2,2,3) and another b with shape (2,1,3), I want a result tensor c with shape (2,1). This tensor would be the result of the following, equivalent python code:
n = a.shape[2]
c = np.zeros((2,n))
for k in range(n):
c += a[:,:,k]*b[:,:,k]
The documentation says that the optional parameter axes:
If an int N, sum over the last N axes of a and the first N axes of b in order. The sizes of the corresponding axes must match.
But I don't understand which "axes" are needed here (furthermore, when axes is a tuple or a tuple of tuples it gets even more confusing). Examples aren't very clear to me either.
The way tensordot works, it won't work here (not at least directly) because of the alignment requirement along the first axes. You can use np.einsum though to solve your case -
c = np.einsum('ijk,ilk->ij',a,b)
Alternatively, use np.matmul/#-operator (Python 3.x) -
np.matmul(a,b.swapaxes(1,2))[...,0] # or (a # b.swapaxes(1,2))[...,0]
I am trying to calculate the first and second order moments for a portfolio of stocks (i.e. expected return and standard deviation).
expected_returns_annual
Out[54]:
ticker
adj_close CNP 0.091859
F -0.007358
GE 0.095399
TSLA 0.204873
WMT -0.000943
dtype: float64
type(expected_returns_annual)
Out[55]: pandas.core.series.Series
weights = np.random.random(num_assets)
weights /= np.sum(weights)
returns = np.dot(expected_returns_annual, weights)
So normally the expected return is calculated by
(x1,...,xn' * (R1,...,Rn)
with x1,...,xn are weights with a constraint that all the weights have to sum up to 1 and ' means that the vector is transposed.
Now I am wondering a bit about the numpy dot function, because
returns = np.dot(expected_returns_annual, weights)
and
returns = np.dot(expected_returns_annual, weights.T)
give the same results.
I tested also the shape of weights.T and weights.
weights.shape
Out[58]: (5,)
weights.T.shape
Out[59]: (5,)
The shape of weights.T should be (,5) and not (5,), but numpy displays them as equal (I also tried np.transpose, but there is the same result)
Does anybody know why numpy behave this way? In my opinion the np.dot product automatically shape the vector the right why so that the vector product work well. Is that correct?
Best regards
Tom
The semantics of np.dot are not great
As Dominique Paul points out, np.dot has very heterogenous behavior depending on the shapes of the inputs. Adding to the confusion, as the OP points out in his question, given that weights is a 1D array, np.array_equal(weights, weights.T) is True (array_equal tests for equality of both value and shape).
Recommendation: use np.matmul or the equivalent # instead
If you are someone just starting out with Numpy, my advice to you would be to ditch np.dot completely. Don't use it in your code at all. Instead, use np.matmul, or the equivalent operator #. The behavior of # is more predictable than that of np.dot, while still being convenient to use. For example, you would get the same dot product for the two 1D arrays you have in your code like so:
returns = expected_returns_annual # weights
You can prove to yourself that this gives the same answer as np.dot with this assert:
assert expected_returns_annual # weights == expected_returns_annual.dot(weights)
Conceptually, # handles this case by promoting the two 1D arrays to appropriate 2D arrays (though the implementation doesn't necessarily do this). For example, if you have x with shape (N,) and y with shape (M,), if you do x # y the shapes will be promoted such that:
x.shape == (1, N)
y.shape == (M, 1)
Complete behavior of matmul/#
Here's what the docs have to say about matmul/# and the shapes of inputs/outputs:
If both arguments are 2-D they are multiplied like conventional matrices.
If either argument is N-D, N > 2, it is treated as a stack of matrices residing in the last two indexes and broadcast accordingly.
If the first argument is 1-D, it is promoted to a matrix by prepending a 1 to its dimensions. After matrix multiplication the prepended 1 is removed.
If the second argument is 1-D, it is promoted to a matrix by appending a 1 to its dimensions. After matrix multiplication the appended 1 is removed.
Notes: the arguments for using # over dot
As hpaulj points out in the comments, np.array_equal(x.dot(y), x # y) for all x and y that are 1D or 2D arrays. So why do I (and why should you) prefer #? I think the best argument for using # is that it helps to improve your code in small but significant ways:
# is explicitly a matrix multiplication operator. x # y will raise an error if y is a scalar, whereas dot will make the assumption that you actually just wanted elementwise multiplication. This can potentially result in a hard-to-localize bug in which dot silently returns a garbage result (I've personally run into that one). Thus, # allows you to be explicit about your own intent for the behavior of a line of code.
Because # is an operator, it has some nice short syntax for coercing various sequence types into arrays, without having to explicitly cast them. For example, [0,1,2] # np.arange(3) is valid syntax.
To be fair, while [0,1,2].dot(arr) is obviously not valid, np.dot([0,1,2], arr) is valid (though more verbose than using #).
When you do need to extend your code to deal with many matrix multiplications instead of just one, the ND cases for # are a conceptually straightforward generalization/vectorization of the lower-D cases.
I had the same question some time ago. It seems that when one of your matrices is one dimensional, then numpy will figure out automatically what you are trying to do.
The documentation for the dot function has a more specific explanation of the logic applied:
If both a and b are 1-D arrays, it is inner product of vectors
(without complex conjugation).
If both a and b are 2-D arrays, it is matrix multiplication, but using
matmul or a # b is preferred.
If either a or b is 0-D (scalar), it is equivalent to multiply and
using numpy.multiply(a, b) or a * b is preferred.
If a is an N-D array and b is a 1-D array, it is a sum product over
the last axis of a and b.
If a is an N-D array and b is an M-D array (where M>=2), it is a sum
product over the last axis of a and the second-to-last axis of b:
In NumPy, a transpose .T reverses the order of dimensions, which means that it doesn't do anything to your one-dimensional array weights.
This is a common source of confusion for people coming from Matlab, in which one-dimensional arrays do not exist. See Transposing a NumPy Array for some earlier discussion of this.
np.dot(x,y) has complicated behavior on higher-dimensional arrays, but its behavior when it's fed two one-dimensional arrays is very simple: it takes the inner product. If we wanted to get the equivalent result as a matrix product of a row and column instead, we'd have to write something like
np.asscalar(x # y[:, np.newaxis])
adding a trailing dimension to y to turn it into a "column", multiplying, and then converting our one-element array back into a scalar. But np.dot(x,y) is much faster and more efficient, so we just use that.
Edit: actually, this was dumb on my part. You can, of course, just write matrix multiplication x # y to get equivalent behavior to np.dot for one-dimensional arrays, as tel's excellent answer points out.
The shape of weights.T should be (,5) and not (5,),
suggests some confusion over the shape attribute. shape is an ordinary Python tuple, i.e. just a set of numbers, one for each dimension of the array. That's analogous to the size of a MATLAB matrix.
(5,) is just the way of displaying a 1 element tuple. The , is required because of older Python history of using () as a simple grouping.
In [22]: tuple([5])
Out[22]: (5,)
Thus the , in (5,) does not have a special numpy meaning, and
In [23]: (,5)
File "<ipython-input-23-08574acbf5a7>", line 1
(,5)
^
SyntaxError: invalid syntax
A key difference between numpy and MATLAB is that arrays can have any number of dimensions (upto 32). MATLAB has a lower boundary of 2.
The result is that a 5 element numpy array can have shapes (5,), (1,5), (5,1), (1,5,1)`, etc.
The handling of a 1d weight array in your example is best explained the np.dot documentation. Describing it as inner product seems clear enough to me. But I'm also happy with the
sum product over the last axis of a and the second-to-last axis of b
description, adjusted for the case where b has only one axis.
(5,) with (5,n) => (n,) # 5 is the common dimension
(n,5) with (5,) => (n,)
(n,5) with (5,1) => (n,1)
In:
(x1,...,xn' * (R1,...,Rn)
are you missing a )?
(x1,...,xn)' * (R1,...,Rn)
And the * means matrix product? Not elementwise product (.* in MATLAB)? (R1,...,Rn) would have size (n,1). (x1,...,xn)' size (1,n). The product (1,1).
By the way, that raises another difference. MATLAB expands dimensions to the right (n,1,1...). numpy expands them to the left (1,1,n) (if needed by broadcasting). The initial dimensions are the outermost ones. That's not as critical a difference as the lower size 2 boundary, but shouldn't be ignored.
I am trying to get rid of the for loop and instead do an array-matrix multiplication to decrease the processing time when the weights array is very large:
import numpy as np
sequence = [np.random.random(10), np.random.random(10), np.random.random(10)]
weights = np.array([[0.1,0.3,0.6],[0.5,0.2,0.3],[0.1,0.8,0.1]])
Cov_matrix = np.matrix(np.cov(sequence))
results = []
for w in weights:
result = np.matrix(w)*Cov_matrix*np.matrix(w).T
results.append(result.A)
Where:
Cov_matrix is a 3x3 matrix
weights is an array of n lenght with n 1x3 matrices in it.
Is there a way to multiply/map weights to Cov_matrix and bypass the for loop? I am not very familiar with all the numpy functions.
I'd like to reiterate what's already been said in another answer: the np.matrix class has much more disadvantages than advantages these days, and I suggest moving to the use of the np.array class alone. Matrix multiplication of arrays can be easily written using the # operator, so the notation is in most cases as elegant as for the matrix class (and arrays don't have several restrictions that matrices do).
With that out of the way, what you need can be done in terms of a call to np.einsum. We need to contract certain indices of three matrices while keeping one index alone in two matrices. That is, we want to perform w_{ij} * Cov_{jk} * w.T_{ki} with a summation over j, k, giving us an array with i indices. The following call to einsum will do:
res = np.einsum('ij,jk,ik->i', weights, Cov_matrix, weights)
Note that the above will give you a single 1d array, whereas you originally had a list of arrays with shape (1,1). I suspect the above result will even make more sense. Also, note that I omitted the transpose in the second weights argument, and this is why the corresponding summation indices appear as ik rather than ki. This should be marginally faster.
To prove that the above gives the same result:
In [8]: results # original
Out[8]: [array([[0.02803215]]), array([[0.02280609]]), array([[0.0318784]])]
In [9]: res # einsum
Out[9]: array([0.02803215, 0.02280609, 0.0318784 ])
The same can be achieved by working with the weights as a matrix and then looking at the diagonal elements of the result. Namely:
np.diag(weights.dot(Cov_matrix).dot(weights.transpose()))
which gives:
array([0.03553664, 0.02394509, 0.03765553])
This does more calculations than necessary (calculates off-diagonals) so maybe someone will suggest a more efficient method.
Note: I'd suggest slowly moving away from np.matrix and instead work with np.array. It takes a bit of getting used to not being able to do A*b but will pay dividends in the long run. Here is a related discussion.
I am solving a linear system of equations Ax=b.
It is known that A is square and of full rank, but it is the result of a few matrix multiplications, say A = numpy.dot(C,numpy.dot(D,E)) in which the result can be 1x1 depending on the inputs C,D,E. In that case A is a float.
b is ensured to be a vector, even when it is a 1x1 one.
I am currently doing
A = numpy.dot(C,numpy.dot(D,E))
try:
x = numpy.linalg.solve(A,b)
except:
x = b[0] / A
I searched numpy's documentation and didn't find other alternatives for solve and dot that would accept scalars for the first or output arrays for the second. Actually numpy.linalg.solve requires dimension at least 2. If we were going to produce an A = numpy.array([5]) it would complain too.
Is there some alternative that I missed?
in which the result can be 1x1 depending on the inputs C,D,E. In that case A is a float.
This is not true, it is a 1x1 matrix, as expected
x=np.array([[1,2]])
z=x.dot(x.T) # 1x2 matrix times 2x1
print(z.shape) # (1, 1)
which works just fine with linalg.solve
linalg.solve(z, z) # returns [[1]], as expected
While you could expand the dimensions of A:
A = numpy.atleast_2d(A)
it sounds like A never should have been a float in the first place, and you should instead fix whatever is causing it to be one.
I have the following line of code in MATLAB which I am trying to convert to Python numpy:
pred = traindata(:,2:257)*beta;
In Python, I have:
pred = traindata[ : , 1:257]*beta
beta is a 256 x 1 array.
In MATLAB,
size(pred) = 1389 x 1
But in Python,
pred.shape = (1389L, 256L)
So, I found out that multiplying by the beta array is producing the difference between the two arrays.
How do I write the original Python line, so that the size of pred is 1389 x 1, like it is in MATLAB when I multiply by my beta array?
I suspect that beta is in fact a 1D numpy array. In numpy, 1D arrays are not row or column vectors where MATLAB clearly makes this distinction. These are simply 1D arrays agnostic of any shape. If you must, you need to manually introduce a new singleton dimension to the beta vector to facilitate the multiplication. On top of this, the * operator actually performs element-wise multiplication. To perform matrix-vector or matrix-matrix multiplication, you must use numpy's dot function to do so.
Therefore, you must do something like this:
import numpy as np # Just in case
pred = np.dot(traindata[:, 1:257], beta[:,None])
beta[:,None] will create a 2D numpy array where the elements from the 1D array are populated along the rows, effectively making a column vector (i.e. 256 x 1). However, if you have already done this on beta, then you don't need to introduce the new singleton dimension. Just use dot normally:
pred = np.dot(traindata[:, 1:257], beta)